Open In App

Replace even nodes of a doubly linked list with the elements of array

Given a doubly linked list and an array with only odd values. Both are of equal size N. The task is replace all node which have even value with the Array elements from left to right.
 

Examples:

Input : List = 6 9 8 7 4 
Arr[] = {3, 5, 23, 17, 1} 
Output : List = 3 9 5 7 23

Input : List = 9 14 7 12 8 13 
Arr[] = {5, 1, 17, 21, 11, 7} 
Output : List = 9 5 7 1 17 13

Approach: The idea is to traverse the nodes of the doubly linked list one by one and get the pointer of the nodes having even data then replace by the value of the array and increment the index of the array and move to the next node in the linked list.

Algorithm:

Below is the implementation of the above approach:




// C++ implementation to create
// odd doubly linked list
#include <bits/stdc++.h>
using namespace std;
 
// Node of the doubly linked list
struct Node {
    int data;
    Node *prev, *next;
};
 
// function to insert a node at the beginning
// of the Doubly Linked List
void push(Node** head_ref, int new_data)
{
    // allocate node
    Node* new_node = (Node*)malloc(sizeof(struct Node));
 
    // put in the data
    new_node->data = new_data;
 
    // since we are adding at the beginning,
    // prev is always NULL
    new_node->prev = NULL;
 
    // link the old list of the new node
    new_node->next = (*head_ref);
 
    // change prev of head node to new node
    if ((*head_ref) != NULL)
        (*head_ref)->prev = new_node;
 
    // move the head to point to the new node
    (*head_ref) = new_node;
}
 
// function to make all node is odd
void makeOddNode(Node** head_ref, int A[], int n)
{
    Node* ptr = *head_ref;
    Node* next;
    int i = 0;
    // traverse list till last node
    while (ptr != NULL) {
        next = ptr->next;
        // check if node is even then
        // replace it and increment in i
        if (ptr->data % 2 == 0) {
            ptr->data = A[i];
            i++;
        }
        ptr = next;
    }
    // return sum of nodes which is divided by K
}
// function to print nodes in a
// given doubly linked list
void printList(Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
}
 
// Driver program to test above
int main()
{
    // start with the empty list
    Node* head = NULL;
 
    // create the doubly linked list
    // 6 <=> 9 <=> 8 <=> 7 <=> 4
    int Arr[] = { 3, 5, 23, 17, 1 };
    push(&head, 4);
    push(&head, 7);
    push(&head, 8);
    push(&head, 9);
    push(&head, 6);
    int n = sizeof(Arr) / sizeof(Arr[0]);
    cout << "Original List: ";
    printList(head);
    cout << endl;
    makeOddNode(&head, Arr, n);
    cout << "New odd List: ";
    printList(head);
}




// Java implementation to create
// odd doubly linked list
class GFG
{
 
// Node of the doubly linked list
static class Node
{
    int data;
    Node prev, next;
};
 
// function to insert a node at the beginning
// of the Doubly Linked List
static Node push(Node head_ref, int new_data)
{
    // allocate node
    Node new_node = new Node();
 
    // put in the data
    new_node.data = new_data;
 
    // since we are adding at the beginning,
    // prev is always null
    new_node.prev = null;
 
    // link the old list of the new node
    new_node.next = (head_ref);
 
    // change prev of head node to new node
    if ((head_ref) != null)
        (head_ref).prev = new_node;
 
    // move the head to point to the new node
    (head_ref) = new_node;
    return head_ref;
}
 
// function to make all node is odd
static Node makeOddNode(Node head_ref, int A[], int n)
{
    Node ptr = head_ref;
    Node next;
    int i = 0;
    // traverse list till last node
    while (ptr != null)
    {
        next = ptr.next;
         
        // check if node is even then
        // replace it and increment in i
        if (ptr.data % 2 == 0)
        {
 
            ptr.data = A[i];
            i++;
        }
        ptr = next;
    }
     
    // return sum of nodes which is divided by K
    return head_ref;
}
 
// function to print nodes in a
// given doubly linked list
static void printList(Node head)
{
    while (head != null)
    {
        System.out.print( head.data + " ");
        head = head.next;
    }
}
 
// Driver code
public static void main(String args[])
{
    // start with the empty list
    Node head = null;
 
    // create the doubly linked list
    // 6 <=> 9 <=> 8 <=> 7 <=> 4
    int Arr[] = { 3, 5, 23, 17, 1 };
    head = push(head, 4);
    head = push(head, 7);
    head = push(head, 8);
    head = push(head, 9);
    head = push(head, 6);
    int n = Arr.length;
    System.out.print( "Original List: ");
    printList(head);
    System.out.println();
    head = makeOddNode(head, Arr, n);
    System.out.print("New odd List: ");
    printList(head);
}
}
 
// This code is contributed by Arnab Kundu




# Python3 implementation to
# create odd doubly linked list
 
# Node of the doubly linked list
class Node:
     
    def __init__(self, data):
        self.data = data
        self.prev = None
        self.next = None
 
# Function to insert a node at the
# beginning of the Doubly Linked List
def push(head_ref, new_data):
 
    # allocate node
    new_node = Node(new_data)
 
    # link the old list of the new node
    new_node.next = head_ref
 
    # change prev of head node to new node
    if head_ref != None:
        head_ref.prev = new_node
 
    # move the head to point to the new node
    head_ref = new_node
    return head_ref
 
# Function to make all node is odd
def makeOddNode(head_ref, A, n):
 
    ptr = head_ref
    i = 0
     
    # traverse list till last node
    while ptr != None:
         
        next = ptr.next
         
        # check if node is even then
        # replace it and increment in i
        if ptr.data % 2 == 0:
            ptr.data = A[i]
            i += 1
         
        ptr = next
     
    # return sum of nodes which is divided by K
 
# Function to print nodes in a
# given doubly linked list
def printList(head):
 
    while head != None:
        print(head.data, end = " ")
        head = head.next
 
# Driver Code
if __name__ == "__main__":
 
    # start with the empty list
    head = None
 
    # create the doubly linked list
    # 6 <=> 9 <=> 8 <=> 7 <=> 4
    Arr = [3, 5, 23, 17, 1]
    head = push(head, 4)
    head = push(head, 7)
    head = push(head, 8)
    head = push(head, 9)
    head = push(head, 6)
    n = len(Arr)
     
    print("Original List:", end = " ")
    printList(head)
    print()
     
    makeOddNode(head, Arr, n)
    print("New odd List:", end = " ")
    printList(head)
 
# This code is contributed by Rituraj Jain




// C# implementation to create
// odd doubly linked list
using System;
     
class GFG
{
 
// Node of the doubly linked list
public class Node
{
    public int data;
    public Node prev, next;
};
 
// function to insert a node at the beginning
// of the Doubly Linked List
static Node push(Node head_ref, int new_data)
{
    // allocate node
    Node new_node = new Node();
 
    // put in the data
    new_node.data = new_data;
 
    // since we are adding at the beginning,
    // prev is always null
    new_node.prev = null;
 
    // link the old list of the new node
    new_node.next = (head_ref);
 
    // change prev of head node to new node
    if ((head_ref) != null)
        (head_ref).prev = new_node;
 
    // move the head to point to the new node
    (head_ref) = new_node;
    return head_ref;
}
 
// function to make all node is odd
static Node makeOddNode(Node head_ref, int []A, int n)
{
    Node ptr = head_ref;
    Node next;
    int i = 0;
     
    // traverse list till last node
    while (ptr != null)
    {
        next = ptr.next;
         
        // check if node is even then
        // replace it and increment in i
        if (ptr.data % 2 == 0)
        {
 
            ptr.data = A[i];
            i++;
        }
        ptr = next;
    }
     
    // return sum of nodes which is divided by K
    return head_ref;
}
 
// function to print nodes in a
// given doubly linked list
static void printList(Node head)
{
    while (head != null)
    {
        Console.Write( head.data + " ");
        head = head.next;
    }
}
 
// Driver code
public static void Main(String []args)
{
    // start with the empty list
    Node head = null;
 
    // create the doubly linked list
    // 6 <=> 9 <=> 8 <=> 7 <=> 4
    int []Arr = { 3, 5, 23, 17, 1 };
    head = push(head, 4);
    head = push(head, 7);
    head = push(head, 8);
    head = push(head, 9);
    head = push(head, 6);
    int n = Arr.Length;
    Console.WriteLine( "Original List: ");
    printList(head);
    Console.WriteLine();
    head = makeOddNode(head, Arr, n);
    Console.WriteLine("New odd List: ");
    printList(head);
}
}
 
// This code contributed by Rajput-Ji




<script>
 
// JavaScript implementation to create
// odd doubly linked list
// Node of the doubly linked list
class Node {
    constructor(val) {
        this.data = val;
        this.prev = null;
        this.next = null;
    }
}
 
    // function to insert a node at the beginning
    // of the Doubly Linked List
    function push(head_ref , new_data) {
        // allocate node
       var new_node = new Node();
 
        // put in the data
        new_node.data = new_data;
 
        // since we are adding at the beginning,
        // prev is always null
        new_node.prev = null;
 
        // link the old list of the new node
        new_node.next = (head_ref);
 
        // change prev of head node to new node
        if ((head_ref) != null)
            (head_ref).prev = new_node;
 
        // move the head to point to the new node
        (head_ref) = new_node;
        return head_ref;
    }
 
    // function to make all node is odd
    function makeOddNode(head_ref , A , n) {
        var ptr = head_ref;
        var next;
        var i = 0;
        // traverse list till last node
        while (ptr != null) {
            next = ptr.next;
 
            // check if node is even then
            // replace it and increment in i
            if (ptr.data % 2 == 0) {
 
                ptr.data = A[i];
                i++;
            }
            ptr = next;
        }
 
        // return sum of nodes which is divided by K
        return head_ref;
    }
 
    // function to print nodes in a
    // given doubly linked list
    function printList(head) {
        while (head != null) {
            document.write(head.data + " ");
            head = head.next;
        }
    }
 
    // Driver code
     
        // start with the empty list
        var head = null;
 
        // create the doubly linked list
        // 6 <=> 9 <=> 8 <=> 7 <=> 4
        var Arr = [ 3, 5, 23, 17, 1 ];
        head = push(head, 4);
        head = push(head, 7);
        head = push(head, 8);
        head = push(head, 9);
        head = push(head, 6);
        var n = Arr.length;
        document.write("Original List: ");
        printList(head);
        document.write("<br/>");
        head = makeOddNode(head, Arr, n);
        document.write("New odd List: ");
        printList(head);
 
// This code contributed by umadevi9616
 
</script>

Output: 
Original List: 6 9 8 7 4 
New odd List: 3 9 5 7 23

 

Time Complexity: O(N), where N is the total number of nodes.
Auxiliary Space: O(1) because it is using constant space
 


Article Tags :